
#1
Apr1206, 01:52 PM

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"The probability that a seed will sprout is 3/4. If 7 seeds are planted, find, to the nearest tenthousandth, the probability that at least 5 will sprout."
The answer is .7564, but I don't know how to get there. I tried (3/4^5) + (3/4^6) + (3/4^7), but that doesn't work. Neither does (3/4*5) + (3/4*6) +(3/4*7)... am I overthinking the question, or is there a formula involved that I don't know? Thanks! 



#2
Apr1206, 01:59 PM

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This is a binomial distribution problem.
You have n seeds. What is the probability that exactly k seeds will sprout? Take into account that the specification of which will sprout and which don't is not important. 



#3
Apr1206, 02:17 PM

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Now, remember that the events of whether a seed sprouts or not is independent of whether the other seeds sprouts.
Also, the probability that a single sprout EITHER sprouts OR not sprouts is obviously: [tex]1=\frac{1}{4}+\frac{3}{4}[/tex] where 1/4 is the probability of nonsprouting, 3/4 the probability of sprouting. Suppose we only have TWO seeds Consider the following identity: [tex]1=(\frac{1}{4}+\frac{3}{4})^{2}=\frac{1}{4}*\frac{1}{4}+2*\frac{1}{4}*\ frac{3}{4}+\frac{3}{4}*\frac{3}{4}[/tex] Now, the first term, 1/4*1/4=1/16 is clearly the probability that neither of the two seeds sprouts, the second term, 2*1/4*3/4=6/16 must be the probability that only one of the seeds sprouts, whereas the last term, 3/4*3/4=9/16 is the probability that BOTH sprout. Considering this, how could you set up a similar scheme for the probabilities for 7 seeds? 



#4
Apr1306, 09:43 AM

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stuck on probability question
arildno, that sounds much more complicated than necessary!
Here's how I would do it: What is the probability that all 7 will sprout? You should know enough to argue that, assuming the sprouting of one seed is independent of the others, you just multiply the probabilities: (3/4)(3/4)(3/4)(3/4)(3/4)(3/4)(3/4)= (3/4)^{7}. What is the probability that exactly 6 will sprout? Well, what is the probability that the first 6 sprout but the last doesn't. Again it is the individual probabilities multiplied together: (3/4)(3/4)(3/4)(3/4)(3/4)(3/4)(1/4)= (3/4)^{6}. But we're not done: what about the first 5 sprout, the sixth doesn't, and the seventh does sprout? Since multiplication is commutative, that will be again exactly (3/4)^{6} and, in fact, for any of the 7!/(6!1!)= 7 different ways that could happen (Taking "S" to mean that seed sprouts and "N" to mean it doesn't, we ould have SSSSSSN, SSSSSNS, SSSSNSS, SSSNSSS, SSNSSSS, SNSSSSS, or NSSSSSS, 7 possible orders) it will be (3/4)[/sup]6[/sup](1/4) for each and so the probability exactly 6 seeds sprout is 7(3/4)[/sup]6[/sup](1/4). What about 5 sprounting and 2 not? Hopefully, you can see now that the probability of a particular order but any order, is (3/4)^{5}(1/4)^{2}. Now how many different orders can we have where 5 sprout and 2 do not? How many different order for the letters SSSSSNN are there? This is where the "binomial coefficient comes in. Once you have found the probability that exactly 7, exactly 6, and exactly 5 seed sprout, add them. 



#5
Apr1306, 01:18 PM

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Why is [itex]1=1^{7}[/itex] a complicated identity?
Okay, I'll admit I have an unreasonable fondness for the expression.. 



#6
Apr1306, 03:40 PM

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#7
Apr1406, 11:42 AM

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ummm... please take into account that I am 16 and stupid




#8
Apr1406, 11:46 AM

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Well, did you understand how we might find the probabilities of events in the case of two seeds?
For example, referring to my earlier post: Of two seeds, what must be the probability that AT LEAST one seed sprouts? 



#9
Apr2106, 12:02 PM

P: 51

this might help
by the binomial distribution we have to consider the prob of either 5 or 6 or 7 sproutings Thus 7C5 (3/4)^5 (1/4)^2 + 7C6 (3/4)^6 (1/4) + 7C7 (3/4)^7 C stands for combination (hope you understand the binomial notation) Cheers Sparsh 


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